Continuous map

First form

The first form:

$$f:A\rightarrow B$$ is continuous at $$a$$ if:
$$\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon$$ (note the implicit $$\forall x\in A$$ )

Second form

Armed with the knowledge of what a metric space is (the notion of distance), you can extend this to the more general:

$$f:(A,d)\rightarrow(B,d')$$ is continuous at $$a$$ if:
$$\forall\epsilon>0\exists\delta>0:d(x,a)<\delta\implies d'(f(x),f(a))<\epsilon$$
$$\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))$$

In both cases the implicit

$$\forall x$$ is present. Basic type inference (the $$B_\epsilon(f(a))$$ is a ball about $$f(a)\in B$$ thus it is a ball in $$B$$ using the metric $$d'$$ )

Third form

The most general form, continuity between topologies

$$f:(A,\mathcal{J})\rightarrow(B,\mathcal{K})$$ is continuous if
$$\forall U\in\mathcal{K}\ f^{-1}(U)\in\mathcal{J}$$ - that is the pre-image of all open sets in $$(A,\mathcal{J})$$ is open.

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TODO: The most important Theorem, that these two continuity definitions are the same

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